A and B worked together to complete a certain task in three hours. If A works alone, he can complete the task in five hours. If B works alone, how much longer will it take him to complete the task?
I have the answer to this question, but I just don't get how that works. Please help me out.
On
We use the fact that speed is inversely proportional to time. Let us denote:
$v_A$ - speed of $A$ working alone,
$v_B$ - speed of $B$ working alone,
$t_A$ - time needed for $A$ to finish the job alone,
$t_B$ - time needed for $B$ to finish the job alone,
$t$ - time needed for $A$ and $B$ to finish the job together.
Then by inverse proportionality we have $$ v_At_A = v_Bt_B = (v_A+v_B)t$$ and when we substitute $t_A = 5$, $t = 3$ we get $$5 v_A = 3v_A + 3v_B\implies v_A = \frac 32 v_B$$ and $$v_Bt_B = \left(\frac 32v_B\right)t_A \implies t_B = \frac 32 t_A\implies t_B = \frac{15}2$$ so $B$ needs to work $7$ and half hours to get the job done.
Assuming that their working speed adds if they work together:
Working Speed of person A = 1/5 task per hour
Working Speed of A + B = 1/3 task per hour
They finish the task in 3 hours, so A has done 3/5 of the task in that time, which means Person B must have done 2/5 of the task.
If we divide the work done by Person B in 3 hours by 3, we get the work Person B does in 1 hour.
(2/5)/3 = 2/15
The working speed of Person B is 2/15 of the task per hour, which means it takes Person B 7.5 hours of work to get the task done.